August 29, 2006 12:08pm by Steve | 
Here’s an old, but still tricky question of probability. You will find it discussed many places on the Net, but give yourself some time to think about it before looking elsewhere. Be warned that looking elsewhere may only lead you to differing conclusions anyway…
Imagine you are a contestant on a game show. You have reached the final round and have to choose one of three closed boxes to claim your prize. There is only one prize, so two of the boxes will be empty. The show’s host knows which box contains the prize, but isn’t telling.
You pick a box. Now, before you open it, the host opens one of the other boxes and shows you that it is empty. He then asks you if you want to switch your choice to the remaining box or stick with your first choice.
What should you do in order to maximize your chances of winning the prize? Stick or switch?
If you want to look this up, it is commonly referred to as the "Monty Hall problem".
For an easier explanation, with a clear example, see the extended entry below.
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(Contains possible spoilers)
My initial knee-jerk was “it doesn’t matter.” I clicked the link, but didn’t read anything remembering you said give yourself time. So I thought about it and realized that I missed an important implied fact…that you told the host what box you chose. For some reason I assumed that while you chose a box, he didn’t know what it was.
In any case, this puts a whole different paint job on things and now I see the conundrum.
I tend to oversimplify things (oddly) so I’m going to say that the fact that there are two boxes keeps the odds at 50/50 that your first choice was right.
Well I just read wikipedia and see I fell into the “typical” mindset. I never liked math.
I have seen this problem before and I know that some so called expert mathematicians insist that your odds are better if you switch. Their reasoin is that with 3 boxes your odds of picking the correct one is 1 in 3. With two boxes it is 1 in 2. Therefore switching boxes gives 1 in 2 odds, not switching leaves with 1 in 3. I do not agree. When you are down to 2 boxes your odds are 1 in 2 no matter what preceeded it.Look at it this way. You pick one box hoping it is the the treasure. Your chances are 1 in 3. Soppose that the box you picked actally has the treasure. Now, the host removes a box that does not have the treasure. Now, with two boxes remaining the one you did not pick has a 1 in 2 probabilty of having the treasure, therefore if you follow the faulty probabilty reasoning you must switch boxes to get a 1 in 2 probabilty, and of course you will end up with the empty box. If you had not switched you would have the correct box. On the other hand… (out of space. See next comment
(Continued from previous comment). On the other hand on your first pick the probability of ending up with the empty box is 2 in 3 and the treasure box has 1 in 3 probability of being chosen. Now if the host removes an empty box, the remaining box has a 1 in 2 probability of being the treasure. Oops, I just convinced myself that you are better off switching! The remaining box has 1 in 2 odds, your first choice has 1 in 3 odds. This works only if we know the box removed is an empty box. If we do not know it, switching would make no difference.